Let $T$ be the set of ordered triples $(x,y,z)$ where $x,y,z$ are integers with $0\leq x,y,z\leq 9$. Players A and b play the following guessing game. Player $A$ chooses a triple $(x,y,z)$ from $T$ and Player B has to find A's triple to win. A move consists of the following: B gives A a triple $(a,b,c)$ and $A$ replies by giving B the number $|x+y-a-b| + |y+z - b-c| + |z + x - c - a|$. Prove that more than 2 moves are required for $B$ to win.
I know that the number of possible values of the pair $(d_i, e_i)$ is $28^2,$ where $d_i$ is the value $A$ gives to $B$ after the ith move. But is it enough to say that just because this value is less than $1000$, the total number of elements in $T$, it is not possible for $B$ to guess $A$'s triple, and if so, why?